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3 years ago

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A child goes down a slide with an initial height of 4m. What is his speed at the bottom of the slide?

1 answer:

iogann1982 [59]3 years ago

3 0

How big is the child

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A rubbit gets down from a rump which its /\x=0.85m in 0.5s, The rubbit's mass is 2kg, what is the net Force?

mestny [16]

Answer: 13.6 N

Explanation:

The equation of motion for the rabbit is:

$\Delta x=V_{ox}t+\frac{1}{2}a_{x}t^{2}$ (1)

Where:

$\Delta x=0.85 m$ is the distance traveled by the rabbit

$V_{ox}=0 m/s$ is the rabbit's initial velocity, assuming it started from rest

$t=0.5 s$ is the time

$a_{x}$ is the acceleration

Isolating $a_{x}$ :

$a_{x}=\frac{2 \Delta x}{t^{2}}$ (2)

$a_{x}=\frac{2 (0.85 m)}{(0.5)^{2}}$ (3)

$a_{x}=6.8 m/s^{2}$ (4)

On the other hand, the force $F_{x}$ is given by:

$F_{x}=m.a_{x}$ (5)

Where $m=2 kg$ is the mass of the rabbit

Substituting (4) in (5):

$F_{x}=(2 kg)(6.8 m/s^{2})$ (6)

Finally:

$F_{x}=13.6 N$

8 0

3 years ago

a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

2 years ago

the deflection angle of the laser beam as it exits the prism is 22. 6º. If the prism had been made of glass instead of polystyre

inessss [21]

Thirty three degrees

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2 years ago

Is baking soda less dense or more dense than water?

boyakko [2]

It’s more dense than air and less dense than liquid!

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3 years ago

If the jet is moving at a speed of 1300 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

Natalka [10]

Answer:

R = 2216m and The normal force of the seat on the pilot is 5008N

Explanation:

See attachment below please.

5 0

3 years ago